$[F(t):F(t^n)]=n$ where $t$ is trascendental

Question

Let $F$ be a field and let $t$ be trascendental over $F$. Prove that $[F(t):F(t^n)]=n$. Obviously $[F(t):F(t^n)]\le n$ since the polynomial $f(x)=x^n-t^n \in F(t^n)$ has $t$ as a root. But I don't know how to prove that it's irreducible over $F(t^n)$. I tried to consider that $t$ is trascendental to deduce some algebraic relations over $F$ on $t$. But it did not work.

Answer 1

Show that $1,t,t^2,\cdots,t^{n-1}$ are linearly independent over $F(t^n)$. (Proof by contradiction: if linearly dependent, then you can find a polynomial over $F$ which has $t$ as a root).

Answer 2

Take an extension of $F$ which contains an $n$ th root of unity, then your claim is clear for this extension, since you have $n$ conjugates. Now deduce your result for $F$ since the degree can only decrease under and extension.

Answer 3

It's not the best approach, but we can do this with Eisenstein's criterion:

If $s=t^n$, then $X^n-s$ is irreducible as a polynomial with coefficients in $F[s]$, because $s$ is prime, the polynomial's non-leading terms are all divisible by $s$, and the constant term is not divisible by $s^2$. By Gauss's lemma, it is irreducible over $F(s)$.